Problem: Complete the square to solve for $x$. $x^{2}-7x+12 = 0$
Move the constant term to the right side of the equation. $x^2 - 7x = -12$ We complete the square by taking half of the coefficient of our $x$ term, squaring it, and adding it to both sides of the equation. The coefficient of our $x$ term is $-7$ , so half of it would be $-\dfrac{7}{2}$ , and squaring it gives us ${\dfrac{49}{4}}$ $x^2 - 7x { + \dfrac{49}{4}} = -12 { + \dfrac{49}{4}}$ We can now rewrite the left side of the equation as a squared term. $( x - \dfrac{7}{2} )^2 = \dfrac{1}{4}$ Take the square root of both sides. $x - \dfrac{7}{2} = \pm\dfrac{1}{2}$ Isolate $x$ to find the solution(s). $x = \dfrac{7}{2}\pm\dfrac{1}{2}$ The solutions are: $x = 4 \text{ or } x = 3$ We already found the completed square: $( x - \dfrac{7}{2} )^2 = \dfrac{1}{4}$